Question: We know that $-\frac{2}{1+4x^2}=-2+8x^2-32x^4+128x^6+...$ for $x\in\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$. Using this fact, find the function that corresponds to the following series. $ -2x+\frac{{8}}{3}x^3-\frac{32}{5}x^5+\frac{128}{7}x^7+...$ Choose 1 answer: Choose 1 answer: (Choice A) A $\ln(-4x)$ (Choice B) B $-\ln(4x)$ (Choice C) C $\ln(4x)$ (Choice D) D $\arctan(-2x)$ (Choice E) E $\arctan(2x)$ (Choice F) F $-\arctan(-2x)$
Explanation: First, note that the derivative of $ -2x+\frac{{8}}{3}x^3-\frac{32}{5}x^5+\frac{128}{7}x^7+...$ is $ -2+8x^2-32x^4+128x^6-...=-\frac{2}{1+4x^2}\,$. Then $\int\big({-2+8x^2-32x^4+128x^6+...}\big)dx=\int{\left(-\frac{2}{1+4x^2}\right)}dx$ We rewrite the integral on the right. $\int{-\frac{2}{1+4x^2}}~dx=\int\frac{-2}{1+(-2x)^2}dx\,$. Now use the $~u$ -substitution $~u=-2x~$ with $~du=-2dx\,$. Then $\begin{aligned}\int{-\frac{2}{1+4x^2}}~dx&=\int\frac{1}{1+u^2}du \\\\ &=\arctan(u)+C \\\\ &=\arctan(-2x)+C \end{aligned}$ Hence, $ -2x+\frac{{8}}{3}x^3-\frac{32}{5}x^5+\frac{128}{7}x^7+...=\arctan(-2x)+C\,$. Now let $x=0\,$ ; since $\arctan(0)=0\,$, we know that $~C=0\,$. Thus, $ -2x+\frac{{8}}{3}x^3-\frac{32}{5}x^5+\frac{128}{7}x^7+...=\arctan(-2x)$